Problem: Simplify and expand the following expression: $ \dfrac{2}{3k + 9}- \dfrac{1}{4k + 36}- \dfrac{2k}{k^2 + 12k + 27} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3k + 9} = \dfrac{2}{3(k + 3)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4k + 36} = \dfrac{1}{4(k + 9)}$ We can factor the quadratic in the third term: $ \dfrac{2k}{k^2 + 12k + 27} = \dfrac{2k}{(k + 3)(k + 9)}$ Now we have: $ \dfrac{2}{3(k + 3)}- \dfrac{1}{4(k + 9)}- \dfrac{2k}{(k + 3)(k + 9)} $ The least common multiple of the denominators is: $ 12(k + 3)(k + 9)$ In order to get the first term over $12(k + 3)(k + 9)$ , multiply by $\dfrac{4(k + 9)}{4(k + 9)}$ $ \dfrac{2}{3(k + 3)} \times \dfrac{4(k + 9)}{4(k + 9)} = \dfrac{8(k + 9)}{12(k + 3)(k + 9)} $ In order to get the second term over $12(k + 3)(k + 9)$ , multiply by $\dfrac{3(k + 3)}{3(k + 3)}$ $ \dfrac{1}{4(k + 9)} \times \dfrac{3(k + 3)}{3(k + 3)} = \dfrac{3(k + 3)}{12(k + 3)(k + 9)} $ In order to get the third term over $12(k + 3)(k + 9)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{2k}{(k + 3)(k + 9)} \times \dfrac{12}{12} = \dfrac{24k}{12(k + 3)(k + 9)} $ Now we have: $ \dfrac{8(k + 9)}{12(k + 3)(k + 9)} - \dfrac{3(k + 3)}{12(k + 3)(k + 9)} - \dfrac{24k}{12(k + 3)(k + 9)} $ $ = \dfrac{ 8(k + 9) - 3(k + 3) - 24k} {12(k + 3)(k + 9)} $ Expand: $ = \dfrac{8k + 72 - 3k - 9 - 24k}{12k^2 + 144k + 324} $ $ = \dfrac{-19k + 63}{12k^2 + 144k + 324}$